PTHH: 2Al + 6HCl → 2AlCl₃ + 3H₂

PTHH: Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O

PTHH: Fe₂O₃ + 6HCl → 2FeCl₃ + 3H₂O

PTHH: K₂O + 2HCl → 2KCl + H₂O

Lần sau em đăng câu hỏi đúng box để mọi người hỗ trợ nha

PTHH: 2Al + 6HCl → 2AlCl₃ + 3H₂

PTHH: Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O

PTHH: Fe₂O₃ + 6HCl → 2FeCl₃ + 3H₂O

PTHH: K₂O + 2HCl → 2KCl + H₂O

Lời giải:

a. ĐKXĐ: $x\leq \frac{1}{5}$

PT $\Leftrightarrow 1-5x=3^2=9$

$\Leftrightarrow 5x=-8\Leftrightarrow x=\frac{-8}{5}$ (tm)

b. ĐKXĐ: $x\geq \frac{3}{5}$

PT $\Leftrightarrow 25x^2-9=4(5x-3)$

$\Leftrightarrow (5x-3)(5x+3)-4(5x-3)=0$

$\Leftrightarrow (5x-3)(5x-1)=0$

$\Leftrightarrow x=\frac{3}{5}$ (tm) hoặc $x=\frac{1}{5}$ (loại)

c. ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow x-4\sqrt{x}+3=0$

$\Leftrightarrow (\sqrt{x}-1)(\sqrt{x}-3)=0$

$\Leftrightarrow \sqrt{x}=1$ hoặc $\sqrt{x}=3$

$\Leftrightarrow x=1$ hoặc $x=9$

d. ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow (\sqrt{x}-2)^2-5=0$

$\Leftrightarrow (\sqrt{x}-2)^2=5$
$\Leftrightarrow \sqrt{x}-2=\pm \sqrt{5}$

$\Leftrightarrow \sqrt{x}=2+\sqrt{5}$ (chọn) hoặc $\sqrt{x}=2-\sqrt{5}$ (loại do âm)

$\Leftrightarrow x=(2+\sqrt{5})^2=9+4\sqrt{5}$

e.ĐKXĐ: $x\geq 3$
PT $\Leftrightarrow 2\sqrt{9}.\sqrt{x-3}-\frac{1}{5}.\sqrt{25}.\sqrt{x-3}-\frac{1}{7}\sqrt{49}.\sqrt{x-3}=20$

$\Leftrightarrow 6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20$

$\Leftrightarrow 4\sqrt{x-3}=20$

$\Leftrightarrow \sqrt{x-3}=5$

$\Leftrightarrow x-3=25$

$\Leftrightarrow x=28$

\(a,ĐK:x\le\dfrac{1}{5}\\ PT\Leftrightarrow1-5x=9\Leftrightarrow x=-\dfrac{8}{5}\\ b,ĐK:x\ge\dfrac{3}{5}\\ PT\Leftrightarrow\sqrt{5x-3}\left(\sqrt{5x+3}-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}5x-3=0\\\sqrt{5x+3}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(tm\right)\\5x+3=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(tm\right)\\x=\dfrac{1}{5}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{3}{5}\)

\(c,ĐK:x\ge0\\ PT\Leftrightarrow2\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\\ d,ĐK:x\ge0\\ PT\Leftrightarrow x-4\sqrt{x}+4-3=0\\ \Leftrightarrow\left(\sqrt{x}-2-\sqrt{3}\right)\left(\sqrt{x}-2+\sqrt{3}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2+\sqrt{3}\\\sqrt{x}=2-\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7+4\sqrt{3}\left(tm\right)\\x=7-4\sqrt{3}\left(tm\right)\end{matrix}\right.\\ e,ĐK:x\ge3\\ PT\Leftrightarrow2\cdot3\sqrt{x-3}-\dfrac{1}{5}\cdot5\sqrt{x-3}-\dfrac{1}{7}\cdot7\sqrt{x-3}=20\\ \Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\\ \Leftrightarrow4\sqrt{x-3}=20\Leftrightarrow\sqrt{x-3}=5\\ \Leftrightarrow x-3=25\Leftrightarrow x=28\left(tm\right)\)

a) \(\sqrt{1-5x}=3\left(đk:x\le\dfrac{1}{5}\right)\)

\(\Leftrightarrow1-5x=9\Leftrightarrow5x=-8\Leftrightarrow x=-\dfrac{8}{5}\left(tm\right)\)

b) \(\sqrt{25x^2-9}=2\sqrt{5x-3}\left(đk:x\ge\dfrac{3}{5}\right)\)

\(\Leftrightarrow25x^2-9=20x-12\)

\(\Leftrightarrow25x^2-20x+3=0\)

\(\Leftrightarrow\left(5x-3\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(tm\right)\\x=\dfrac{1}{5}\left(ktm\right)\end{matrix}\right.\)

c) \(ĐK:x\ge0\)

\(\Leftrightarrow2\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=9\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)

d) \(ĐK:x\ge0\)

\(\Leftrightarrow\left(\sqrt{x}-2-\sqrt{5}\right)\left(\sqrt{x}-2+\sqrt{5}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2+\sqrt{5}\\\sqrt{x}=2-\sqrt{5}\left(VLý\right)\end{matrix}\right.\)

\(\Leftrightarrow x=9+4\sqrt{5}\left(tm\right)\)

e) \(ĐK:x\ge3\)

\(\Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)

\(\Leftrightarrow4\sqrt{x-3}=20\Leftrightarrow\sqrt{x-3}=5\)

\(\Leftrightarrow x-3=25\Leftrightarrow x=28\left(tm\right)\)

\(1,\\ a,=\dfrac{\left(3+2\sqrt{3}\right)\sqrt{3}}{3}+\dfrac{\left(2+\sqrt{2}\right)\left(\sqrt{2}-1\right)}{1}\\ =\dfrac{3\sqrt{3}+6}{3}+\sqrt{2}=\sqrt{3}+1+\sqrt{2}\\ b,=\left(\dfrac{\sqrt{5}+\sqrt{2}}{3}-\dfrac{\sqrt{5}-\sqrt{2}}{3}+1\right)\cdot\dfrac{1}{\left(\sqrt{2}+1\right)^2}\\ =\dfrac{\sqrt{5}+\sqrt{2}-\sqrt{5}+\sqrt{2}+3}{3}\cdot\dfrac{1}{\left(\sqrt{2}+1\right)^2}\\ =\dfrac{2\sqrt{2}+3}{3\left(3+2\sqrt{2}\right)}=\dfrac{1}{3}\)

\(2,\\ A=2x+\sqrt{\left(x-3\right)^2}=2x+\left|x-3\right|\\ =2\left(-5\right)+\left|-5-3\right|=-10+8=-2\\ B=\dfrac{\sqrt{\left(2x+1\right)^2}}{\left(x-4\right)\left(x+4\right)}\left(x-4\right)^2=\dfrac{\left|2x+1\right|\left(x-4\right)}{x+4}\\ B=\dfrac{17\cdot4}{12}=\dfrac{17}{3}\)

1 hadn't seen - were

2 doesn't like - being phoned

3 is eaten

4 is leaving - will wait

5 won't be - will you go

6 is coming - don't want

7 hasn't read - have gone

8 haven't seen - left

9 was washing - was writing

10 sets - goes

11 were talking - walked

12 has stayed - studied - was

13 was made

14 had told - went away

15 will be held

16 had gone - sat - rested

17 watched - had done

B(NGHĨ.LÀ.THẾ)

Xét tam giác ABC có: góc A+góc B+góc C=180^o

=>Góc B+góc C=180^o-góc A=180^o-60^o=120^o

Tổng tia phân giác của góc B và góc C là (góc B)/2+(góc C)/2

=(góc B+góc C)/2=120^o/2=60^o=>góc IBC+góc ICB=60^o

Xét tam giác BIC có: góc IBC+góc ICB+góc BIC=180^o

=>Góc BIC=180^o-(góc IBC+góc ICB)=180^o-60^o=120^o

Vậy góc BIC=60^o

Góc BIC+60^o=>chọn D

Bài nào ạ?

ko có đề bài sao làm

I)1.B

2.B

3.C

4.D

5.C

II)

1.B

2.D

3.B

4.B

5.A

6.B

7.tell....way.

8.C

III)

1.isn't teaching

2.drive

3.has

làm phần nào?

I

1.B 2.B 3B 4D 4.C

II.

1B 2.D 3.B 4.B 5.A 6.B 7.tell - way 8.C

III

1.isn't teaching

2.drive

3.has